Proof $(n^2 + 1)$ in $\Theta(n^3 - 2n - 3)$

Question

I trying resolve this problem, but the first equation have 3 terms and the second equation have 2 terms. I don't know how to resolve this problem. Any idea?

Answer 1

Note that $\Theta(n^3 - 2n - 3)$ is the set of functions $f(n)$ such that $\frac{f(n)}{n^3 - 2n - 3}$ is bounded away from both $0$ and $\infty$ for arbitrarily large $n$.

Since $\frac{n^2+1}{n^3 - 2n - 3} \to 0$ as $n \to \infty$, $n^2+1 \notin \Theta(n^3 - 2n - 3)$.

However, $n^2+1 \in O(n^3 - 2n - 3)$. This is often written $n^2+1 = O(n^3 - 2n - 3)$.

Also note that $\Theta(n^3 - 2n - 3) = \Theta(n^3)$.